Here, Carefully look at the graph.
When it is on x=10, it is approximately 10, (slightly less than 10)
Closest value would be 90, so y/x = 90/10 = 9
So, the density of the graph would be 9 g/cm³
In short, Your Answer would be Option D
Hope this helps!
29.213 cm3
First, calculate the mass of the water used. You do this by subtracting the original mass of the flask from the combined mass of the water and flask, giving:60.735 g - 31.601 g = 29.134 g
So we now know we have 29.134 g of water. To calculate the volume of the flask, simply divide by the density of the water, giving:29.134 g / (0.9973 g/cm3) = 29.213 cm3
Answer : The final pressure in the two containers is, 2.62 atm
Explanation :
Boyle's Law : It is defined as the pressure of the gas is inversely proportional to the volume of the gas at constant temperature and number of moles.

Thus, the expression for final pressure in the two containers will be:


where,
= pressure of N₂ gas = 4.45 atm
= pressure of Ar gas = 2.75 atm
= volume of N₂ gas = 3.00 L
= volume of Ar gas = 2.00 L
P = final pressure of gas = ?
V = final volume of gas = (4.45 + 2.75) L = 7.2 L
Now put all the given values in the above equation, we get:


Thus, the final pressure in the two containers is, 2.62 atm
molecules of water are never destroyed - they go through various uses in a cycle of re-use. beginning in the ocean. a water molecue is attached to the wet suit of a deep sea diver. when the diver gets back on his boat, the water molecule leaves the ocean. Diver dry his suit under the sun. The water molecule is evaporated to the air. It meets up with more water molecules to form cloud. Cloud becomes rain over ground. Rain drains into stream which merges into river. River runs out to the ocean and the water cycle starts anew.
Answer:
5. -24 m/s²
Explanation:
Acceleration: This can be defined as the rate of change of velocity.
The S.I unit of acceleration is m/s².
mathematically,
a = dv/dt ............................ Equation 1
Where a = acceleration, dv/dt = is the differentiation of velocity with respect to time.
But
v = dx(t)/dt
Where,
x(t) = 27t-4.0t³...................... Equation 2
Therefore, differentiating equation 2 with respect to time.
v = dx(t)/dt = 27-12t²............. Equation 3.
Also differentiating equation 3 with respect to time,
a = dv/dt = -24t
a = -24t .................... Equation 4
from the question,
At the end of 1.0 s,
a = -24(1)
a = -24 m/s².
Thus the acceleration = -24 m/s²
The right option is 5. -24 m/s²